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# CAT 2020 Quant Question [Slot 3] with Solution 11

Question

Let m and n be positive integers, If ${x^2} + mx + 2n = 0$ and ${x^2} + 2nx + m = 0$ have real roots, then the smallest possible value of $m + n$ is

1. 7
2. 8
3. 5
4. 6
Option: 4
Solution:

Since the roots are real ${m^2} - 8n \ge 0$ and ${(2n)^2} - 4m \ge 0$ $\Rightarrow {n^2} - m \ge 0$

$\Rightarrow {n^4} \ge {m^2} \ge 8n$

$\Rightarrow n \ge 2$ and $m \ge 4$

Hence the least value of $m + n = 2 + 4 = 6$

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## CAT 2020 Quant questions with Solutions

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